Trắc nghiệm Phương sai và độ lệch chuẩn có đáp án (Nhận biết)

 \(\overline{x}=\frac{{n}_{1}{x}_{1}+{n}_{2}{x}_{2}+\mathrm{...}+{n}_{k}{x}_{k}}{{n}^{2}}\)

 \(\overline{x}=\frac{{n}_{1}{x}_{1}+{n}_{2}{x}_{2}+\mathrm{...}+{n}_{k}{x}_{k}}{n}\)

 \(\overline{x}={n}_{1}{x}_{1}+{n}_{2}{x}_{2}+\mathrm{...}+{n}_{k}{x}_{k}\)

 \(\overline{x}=n\left({n}_{1}{x}_{1}+{n}_{2}{x}_{2}+\mathrm{...}+{n}_{k}{x}_{k}\right)\)

 \({S}_{X}^{2}=\frac{1}{n}\left[{n}_{1}{\left({x}_{1}-\overline{x}\right)}^{2}+{n}_{2}{\left({x}_{2}-\overline{x}\right)}^{2}+\mathrm{...}+{n}_{k}{\left({x}_{k}-\overline{x}\right)}^{2}\right]\)

 \({S}_{X}^{2}=\frac{1}{n}\left[{\left({x}_{1}-\overline{x}\right)}^{2}+{\left({x}_{2}-\overline{x}\right)}^{2}+\mathrm{...}+{\left({x}_{k}-\overline{x}\right)}^{2}\right]\)

 \({S}_{X}^{2}={\left({x}_{1}-\overline{x}\right)}^{2}+{\left({x}_{2}-\overline{x}\right)}^{2}+\mathrm{...}+{\left({x}_{k}-\overline{x}\right)}^{2}\)

 \({S}_{X}^{2}=\frac{1}{{n}^{2}}\left[{n}_{1}{\left({x}_{1}-\overline{x}\right)}^{2}+{n}_{2}{\left({x}_{2}-\overline{x}\right)}^{2}+\mathrm{...}+{n}_{k}{\left({x}_{k}-\overline{x}\right)}^{2}\right]\)

 \({S}_{X}^{2}=\frac{1}{{n}^{2}}\left[{n}_{1}{\left({x}_{1}-\overline{x}\right)}^{2}+{n}_{2}{\left({x}_{2}-\overline{x}\right)}^{2}+\mathrm{...}+{n}_{k}{\left({x}_{k}-\overline{x}\right)}^{2}\right]\)

 \({S}_{X}=\frac{1}{n}\left[{n}_{1}{\left({x}_{1}-\overline{x}\right)}^{2}+{n}_{2}{\left({x}_{2}-\overline{x}\right)}^{2}+\mathrm{...}+{n}_{k}{\left({x}_{k}-\overline{x}\right)}^{2}\right]\)

 \({S}_{X}=\sqrt{\frac{1}{n}\left[{n}_{1}{\left({x}_{1}-\overline{x}\right)}^{2}+{n}_{2}{\left({x}_{2}-\overline{x}\right)}^{2}+\mathrm{...}+{n}_{k}{\left({x}_{k}-\overline{x}\right)}^{2}\right]}\)

 \({S}_{X}^{2}=\frac{1}{n}\sqrt{{n}_{1}{\left({x}_{1}-\overline{x}\right)}^{2}+{n}_{2}{\left({x}_{2}-\overline{x}\right)}^{2}+\mathrm{...}+{n}_{k}{\left({x}_{k}-\overline{x}\right)}^{2}}\)

Tần số lớn nhất. 

Giá trị có tần số lớn nhất. 

Tần số nhỏ nhất. 

Giá trị lớn nhất.